- #1
PLuz
- 60
- 0
I've been working on a problem where I have to find the geodesics for a given Riemannian Manifold. To present my doubt, I tried to find a simpler example that would demonstrate my uncertainty but the one I found, and shall present bellow, has actually a simplification that my problem doesn't, so please ignore that simplification (I shall indicate it when it comes up).
Given a Riemannian manifold with metric
$$\tag{1} ds^{2}=\frac{dr^2}{a+r^2}+r^2\left(d \theta^2+\sin^2(\theta) \, d\phi^{2}\right),$$
consider a curve c(\lambda)=(r(\lambda),\phi(\lambda),\pi/2) whose tangent vector is c'(\lambda)=(r'(\lambda),\phi'(\lambda),0). The geodesic equations (if I didn't mess up) are given by:
\begin{cases}
\tag{2} r''-\frac{r'^{2}}{2(a+r)}-r(a+r)\phi'^{2}=0,\\
\phi''+2\frac{r'}{r}\phi'=0.
\end{cases}
The second equation of Eq.(2) can be integrated, such that:
$$\tag{3} \phi'(\lambda)=\frac{C}{r^{2}(\lambda)},$$
where C is a constant of integration.
Now I introduce a new equation:
$$\tag{4}r'^{2}\left(\frac{1}{a+r^{2}}\right)+\phi'^{2}r^{2}=1,$$
which is basically impose unit speed. Substituting Eq.(3) in Eq.(4) we have that:
$$\tag{5} r'=\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}.$$
(Here is the difference from my case since Eq.(5) can be integrated analytically and in my case, the congener equation can't).
Dividing Eq.(3) by Eq.(5) we have that
$$\tag{6} \frac{d\phi}{dr}=\frac{C}{r^{2}(\lambda)}\frac{1}{\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}}.$$
So my question is:Does Eq.(6) allow me to write the curve c as c(r)=(r,\phi(r)), where \phi(r) is given by (6) and c is a geodesic?
Note: This treatment is based on the Clairaut parametrization but I'm not sure if I can do it in this kind of problem where the g_{rr} component of the metric varies...
Given a Riemannian manifold with metric
$$\tag{1} ds^{2}=\frac{dr^2}{a+r^2}+r^2\left(d \theta^2+\sin^2(\theta) \, d\phi^{2}\right),$$
consider a curve c(\lambda)=(r(\lambda),\phi(\lambda),\pi/2) whose tangent vector is c'(\lambda)=(r'(\lambda),\phi'(\lambda),0). The geodesic equations (if I didn't mess up) are given by:
\begin{cases}
\tag{2} r''-\frac{r'^{2}}{2(a+r)}-r(a+r)\phi'^{2}=0,\\
\phi''+2\frac{r'}{r}\phi'=0.
\end{cases}
The second equation of Eq.(2) can be integrated, such that:
$$\tag{3} \phi'(\lambda)=\frac{C}{r^{2}(\lambda)},$$
where C is a constant of integration.
Now I introduce a new equation:
$$\tag{4}r'^{2}\left(\frac{1}{a+r^{2}}\right)+\phi'^{2}r^{2}=1,$$
which is basically impose unit speed. Substituting Eq.(3) in Eq.(4) we have that:
$$\tag{5} r'=\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}.$$
(Here is the difference from my case since Eq.(5) can be integrated analytically and in my case, the congener equation can't).
Dividing Eq.(3) by Eq.(5) we have that
$$\tag{6} \frac{d\phi}{dr}=\frac{C}{r^{2}(\lambda)}\frac{1}{\sqrt{(a+r^{2})\left(1-\frac{C^{2}}{r^{2}}\right)}}.$$
So my question is:Does Eq.(6) allow me to write the curve c as c(r)=(r,\phi(r)), where \phi(r) is given by (6) and c is a geodesic?
Note: This treatment is based on the Clairaut parametrization but I'm not sure if I can do it in this kind of problem where the g_{rr} component of the metric varies...
Last edited:
